Question

evaluate log2 ^150 interms of x and y such that logo2 ^3=x and logo 2^5=y​

Answer 1

$\textbf{Formula used:}$

$\textbf{Product rule:}$

$\boxed{\bf\,log_a(MN)=log_aM+log_aN}$

$\textbf{Power rule:}$

$\boxed{\bf\,log_aM^n=n\;log_aM}$

$\textbf{Given:}$

$\log_23=x\;\text{and}\;\log_25=y$

$\textbf{To find:}$

$\log_2150\;\text{interms of x and y}$

$\text{Now,}$

$\log_2150$

$=\log_2(2{\times}3{\times}5^2)$

$\text{Using product rule, we get}$

$=\log_22+\log_23+\log_25^2$

$\text{Using power rule, we get}$

$=\log_22+\log_23+2\;\log_25$

$=1+x+2\;y$

$\therefore\boxed{\bf\log_2150=1+x+2\;y}$

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Answer 2

Given :

The log term as $Log_2$ 150

And $Log_23$ = x

      $Log_2$5 = y

To Find :

Evaluate  $Log_2$ 150 in terms of x and y

Solution :

∵  $Log_2$ 150 =  $Log_2$ ( 2 × 3 × 5²)

               [ ∵ Log ( m ×n ) = Log m + Log n ]

                  =  $Log_2$ 2 +  $Log_2$ 3 +  $Log_2$ 5²  

                [ ∵ Log $a^{b}$ = b Log a ]

                  =  $Log_2$ 2 +  $Log_2$ 3 + 2 $Log_2$ 5  

                 =  1 +  $Log_2$ 3 + 2 $Log_2$ 5              [ ∵ $Log_a$ a = 1 ]

                  = 1 + x + y                              [ substituting value of $Log_23$ , $Log_2$5 ]

∴  $Log_2$ 150 =  1 + x + y

Hence, The value of $Log_2$ 150 in terms of x and y is  1 + x + y  Answer